I’m in Melbourne right now, where I recently attended the Hyamfest and the preceding workshop. There were many excellent talks at both the workshop and the conference (more on that in another post), but one thing that I found very interesting is that both Michel Boileau and Cameron Gordon gave talks on the relationships between taut foliations, left-orderable groups, and L-spaces. I haven’t thought seriously about taut foliations in almost ten years, but the subject has been revitalized by its relationship to the theory of Heegaard Floer homology. The relationship tends to be one-way: the existence of a taut foliation on a manifold implies that the Heegard Floer homology of is nontrivial. It would be very interesting if Heegaard Floer homology could be used to decide whether a given manifold admits a taut foliation or not, but for the moment this seems to be out of reach.

Anyway, both Michel and Cameron made use of the (by now 20 year old) classification of taut foliations on Seifert fibered 3-manifolds. The last step of this classification concerns the case when the base orbifold is a sphere; the precise answer was formulated in terms of a conjecture by Jankins and Neumann, proved by Naimi, about rotation numbers. I am ashamed to say that I never actually read Naimi’s argument, although it is not long. The point of this post is to give a new, short, combinatorial proof of the conjecture which I think is “conceptual” enough to digest easily.

The conjecture concerns rotation numbers of circle homeomorphisms. Given an orientation-preserving homeomorphism of the circle , Poincaré defined the so-called rotation number of as follows. Lift to a homeomorphism of the line, then define . Then define to be the reduction of mod .

In fact, the conjecture is about the real-valued rotation numbers of the lifts, and can be stated in the form of a question. Given homeomorphisms of the circle, and lifts to homeomorphisms of the line with (real-valued) rotation numbers , what is the maximum (real-valued) rotation number of the product ? We denote this maximum as . For elementary reasons it satisfies for any integers so it suffices to restrict attention to . It is also elementary to show that is monotone nondecreasing (though not continuous) in both and ; from the form of the answer it suffices to determine for rational.

In this language, what Jankins and Neumann conjectured, and Naimi proved, is the following:

Theorem: where the maximum is taken over all rational and .

We show how to turn this into a combinatorial problem, which can then be solved directly. Given homeomorphisms of the circle, with rotation numbers and respectively, we can choose periodic orbits for and for , so that and , indices taken mod and respectively. Denote the union of the by .

Now, in place of homeomorphisms and consider (discontinuous) maps and defined by for , and similarly for . The point is that we can adjust the dynamics of and on the complement of the and the respectively without changing their rotation number. Replacing and with new satisfying and for all gives . If successive elements of are at least apart, then providing for (and similarly for ) the powers of and have orbits that stay a bounded distance apart.

So in order to find it suffices to study the rotation numbers of as above. Evidently, these rotation numbers depend (in a simple way, which we will now describe) only on the circular order of the points . We encode the circular order of the by a cyclic word of ‘s and ‘s, one for each , and one for each . We define a dynamical system, whose states are the letters of . The transformation acts by moving to the right ‘s (including the we start on, if we start on an ) and the transformation acts by moving to the right ‘s (including the we start on, if we start on a ). Any with ‘s and ‘s is said to be admissible for . For each admissible the transformation acting on has an obvious rotation number, and is the maximum of this rotation number over all admissible . We illustrate this with an example:

Example: To compute the admissible words are (up to cyclic permutation) , , and . Starting on the last (cyclic) letter, and successively applying gives in the first case a rotation number of , in the second case a rotation number of , and in the third case a rotation number of , so .

With this setup established, we now prove the theorem:

Proof: We prove the desired inequality for rational and . Suppose is an admissible word, for which has rotation number , and suppose this is maximal over all , so that . We can decompose (up to cyclic permutation) into subwords so that if denotes the last letter of , then , indices taken mod . We can similarly decompose a cyclic permutation of into subwords so that , indices taken mod . We can choose indices so that and . Let be the subdivision of generated by the and subdivisions. By the definition of the transformations and , each is a letter , and each is a letter , so the endpoints are distinct, and the subdivision has exactly elements. We may permute the letters within each without changing the dynamics, providing we keep the last letter fixed. So we can assume that each is either of the form (if for some ) or of the form (if for some ).

Now suppose some is entirely contained in some . Hence coincides with some and therefore . We claim we can move the string to the left, past the rightmost string of ‘s in (note that ). Since ignores ‘s, we will still have after this transformation. Moreover, since each interval contains the same or fewer ‘s after this move, we have after this transformation; i.e. for the new word we obtain, the rotation number of is no smaller than it was for . So without loss of generality, if is entirely contained in some then we can assume that consists entirely of ‘s; similarly, any contained entirely in can be assumed to consist entirely of ‘s. But this means that contains at most consecutive strings of ‘s and ‘s, and therefore exactly (since each is an and each is a ), so each is of the form . This implies that the and alternate, so that there is a fixed so that for each . Now, so . Since this is true for every , and since , we get an inequality . Similarly, we have an inequality . But , as one can see by considering the dynamics of and on the word . qed

This combinatorial language turns out to be quite flexible, and one can push the techniques substantially further; Alden Walker and I are busy writing this up at the moment. One of the nice aspects of this story is that it gives rise to attractive pictures; the graph of for is the “ziggurat” appearing in the following figure. The vertical faces of the ziggurat correspond to places where is not continuous as a function of .

I should definitely put a better picture here at some stage. Let me explain in words for now: the r and s axes are “horizontal” in the picture, and R is “vertical”. Now, it turns out that R(r,s)=1 for r+s<1; those values of R should be represented by a "flat" triangle at the front of the figure (which has been omitted). The straight (horizontal) line at the front is r+s=1; note that R(p/q,(q-p)/q) = 1+1/q, so there is a vertical line of height 1/q at each point (p/q,(q-p)/q) (these end at the vertices of the "cubes"). There is an order 3 symmetry in the figure, coming from the order 3 symmetry of F_2 interchanging a,b,AB.